SATHEE: Circle 3 Question 14

Circle 3 Question 14

14. A common tangent of the two circles is

(a) $x = 4$

(b) $y = 2$

(d) $x + 2 \sqrt{2} y = 6$

Passage 2

$A$ circle $C$ of radius 1 is inscribed in an equilateral $△ P Q R$ . The points of contact of $C$ with the sides $P Q, Q R$ , $R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation $\sqrt{3} x + y - 6 = 0$ and the point $D$ is $\frac{3 \sqrt{3}}{2}, \frac{3}{2}$ .

Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$ .

$(2008, 12 M)$

Show Answer

Answer:

Correct Answer: 14. (d)

Solution:

Here, equation of common tangent be

$y = m x \pm 2 \sqrt{1 + m^{2}}$

which is also the tangent to

$\begin{aligned} (x - 3)^{2} + y^{2} = 1 \\ \Rightarrow \frac{| 3 m - 0 + 2 \sqrt{1 + m^{2}} |}{\sqrt{m^{2} + 1}} = 1 \\ \Rightarrow 3 m + 2 \sqrt{1 + m^{2}} = \pm \sqrt{1 + m^{2}} \\ \Rightarrow 3 m = - 3 \sqrt{1 + m^{2}} \\ or 3 m = - \sqrt{1 + m^{2}} \\ \Rightarrow m^{2} = 1 + m^{2} or 9 m^{2} = 1 + m^{2} \\ \Rightarrow m \in φ or m = \pm \frac{1}{2 \sqrt{2}} \\ ∴ y = \pm \frac{1}{2 \sqrt{2}} x \pm 2 \sqrt{1 + \frac{1}{8}} \\ \Rightarrow y = \pm \frac{x}{2 \sqrt{2}} \pm \frac{6}{2 \sqrt{2}} \\ \Rightarrow 2 \sqrt{2} y = \pm (x + 6) \\ ∴ x + 2 \sqrt{2} y = 6 \end{aligned}$

Circle 3 Question 14

14. A common tangent of the two circles is

Passage 2

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Circle 3 Question 14

14. A common tangent of the two circles is

Passage 2

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu